Question

intergral of Sec^3(2x)dx

Answer 1

oh this integral lol
let's see if i remember how to do it
i think you start with\[u=\sec(2x)\\dv=\sec^2(2x)\]

Answer 2

Yes that is how I started it.

Answer 3

I got to this part
1/2sec(2x)tan(x)-intergral of tan^2(x)sec(2x)dx

Answer 4

let me do it on my whiteboard one sec

Answer 5

Ok thanks.

Answer 6

looks okay so far, except in the tan the argument should be 2x

Answer 7

Ok, this is where I am stuck at....

Answer 8

use the identity\[\tan^2x=\sec^2x-1\]

Answer 9

O you have to set them = to each other and solve?

Answer 10

it is a repeating integral, if that's what you mean, yes
like \(\int e^x\sin xdx\)

Answer 11

i'm not sure i understood your comment tho :P

Answer 12

Yes that is what I ment.

Answer 13

So this is the answer i got.
(1/2sec(2x)tan(2x)-ln(sec(2x)+tan(2x)))/2+C

Answer 14

looks legit, let me dbl check

Answer 15

Thanks for all your help.

Answer 16

i think it is right but i am being distracted sorry
one sec ...

Answer 17

Its ok, no rush

Answer 18

yep that's what i got :)
sorry that took a while, multitasking

Answer 19

Its ok, you were awesome. Thanks!

Answer 20

welcome!